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Proof step 13 from proof keys-same-fs (previous step)

keys(fs0, ax) = keys(fs1, ax), fs0 | ax.toSet = fs1 | ax.toSet, ∀ adr0. (adr0 ∈ fs0 | (ax.toSet ++ adr) ↔ adr0 ∈ fs1 | (ax.toSet ++ adr)) ∧ (fs0 | (ax.toSet ++ adr))[adr0] = (fs1 | (ax.toSet ++ adr))[adr0] ⊦ fs0[adr].key = fs1[adr].key

User applied rule all left

And obtained the following new goals

Proof step 14:
adr ∈ fs0 | (ax.toSet ++ adr) ↔ adr ∈ fs1 | (ax.toSet ++ adr), (fs0 | (ax.toSet ++ adr))[adr] = (fs1 | (ax.toSet ++ adr))[adr], keys(fs0, ax) = keys(fs1, ax), fs0 | ax.toSet = fs1 | ax.toSet ⊦ fs0[adr].key = fs1[adr].key

Rule argument

pos: left 3 terms: adr abort: F discard: T computed: T

Additional information

Used terms: adr